Old Math 10 Exams David M. McClendon Department of Mathematics Ferris State University 1
Contents Contents Contents 1 General comments on these exams 3 Exams from Fall 016 4.1 Fall 016 Exam 1............................... 4. Fall 016 Exam............................... 8.3 Fall 016 Exam 3............................... 14.4 Fall 016 Final Exam............................ 19 3 Exams from Fall 017 9 3.1 Fall 017 Exam 1............................... 9 3. Fall 017 Exam............................... 33 3.3 Fall 017 Exam 3............................... 38 3.4 Fall 017 Final Exam............................ 4
Chapter 1 General comments on these exams These are the exams I have given in a Math 10 (trigonometry) course at Ferris State University in Fall 016 and Fall 017. Each exam has two sections: on the first section, calculators are not allowed and exact answers must be given, but on the second section, calculators are permitted and decimal approximations are acceptable. Each exam, other than the Fall 017 final exam, is followed by solutions (there may be minor errors in the solutions). These exams correspond to the material in my lecture notes and the 7 th edition of the McKeague textbook as follows: Exam 1: chapters 1 & of my lecture notes / sections A.1, 1.1-1., 3., 3.4-3.5 of the McKeague text Exam : chapter 3 of my lecture notes / sections 1.3-1.5,.1-.4, 3.1, 3.3, 7.1-7.4 of the McKeague text Exam 3: chapters 4 & 5 of my lecture notes / sections 1.3-.5, 7.5-7.6 of the McKeague text Final Exam: chapters 1-7 of my lecture notes / chapters 1-5 and 7 of the McKeague text 3
Chapter Exams from Fall 016.1 Fall 016 Exam 1 No calculator allowed on these problems 1. Convert the following angles from radians to degrees, and draw them in standard position: a) 5π 6 b) π c) 8π 3 d) 4π 4. Let cat be the function defined by cat x = x + 1. Compute each quantity: a) cat + 3 b) cat 3 c) cat() + 3 d) cat 3 Calculators allowed on the rest of the exam 3. Parts (a), (b), (c) and (d) of this question are not related to one another. a) Find an angle measuring between 0 and 360 which is coterminal with 4375. b) Convert to radians (write your answer as a decimal, rounded to two or more decimal places). c) Find the coordinates of a point on the unit circle which is on the terminal side of a 540 angle, drawn in standard position. d) Draw a picture of a triangle which is isoceles, but not equilateral. 4. Parts (a), (b), and (c) of this question are not related to one another. a) Find the distance between the points (, 8) and (5, 3) (write your answer either as an exact answer or as a decimal, rounded to two or more decimal places). 4
.1. Fall 016 Exam 1 b) An angle measures 18 more than its complement. What is the measure of the angle? c) The three angles of a triangle measure x, x + 5 and x 40. Is this triangle an acute triangle, a right triangle, or an obtuse triangle? 5. Parts (a) and (b) of this question are not related to one another. a) A Ferris wheel makes.065 revolutions in a minute. If the radius of the Ferris wheel is 10 feet, what is the linear velocity of someone sitting in a bucket on the edge of the Ferris wheel? b) Find the measure of each angle in this picture: 3x - 3 8x + 38 6. In each picture, find x: 83 5 a) x b) (in this picture, assume the horizontal lines are parallel) 3x - 3 x + 46 x 150 x c) 15 5
.1. Fall 016 Exam 1 Exam 1 Solutions 1. a) 5π 6 = 5 π 6 = 5 30 = 150. In standard position, this angle should point just north of the negative x-axis. b) π = 90. In standard position, this angle points due north. c) 8π 3 = 8 π 3 = 8 60 = 480. In standard position, this angle goes all the way around once, then halfway around again to end up on the negative x-axis. d) 4π 4 = π = 180. In standard position, this angle is on the negative x-axis.. a) cat + 3 = cat() + 3 = ( + 1) + 3 = 5 + 3 = 8. b) cat 3 = cat 6 = 6 + 1 = 37. c) cat() + 3 = ( + 1) + 3 = 5 + 3 = 8. d) cat 3 = (3 + 1) = (10) = 0. 3. a) 4375 360 = 1.15 so the angle is 4375 1 360 = 55. b) π 180 = 3.87 radians. c) 540 = 360 + 180 so the terminal side of this angle points on the negative x-axis. The point on the negative x-axis which is on the unit circle is ( 1, 0). d) The triangle should have two sides that are the same length, but not all three sides of the same length. 4. a) D = (x x 1 ) + (y y 1 ) = (5 ) + ( 3 8) = 3 + ( 11) = 9 + 11 = 130 11.4. b) Let x be the angle; we have x = 18 +(90 x). Solve for x to get x = 54. c) The three angles sum to 180, so x + (x + 5 ) + (x 40 ) = 180. Solve for x to get x = 65. This makes the three angles x = 65, x + 5 = 90 and x 40 = 5. Since one of the angles is 90, the triangle is a right triangle. 5. a) The angular velocity is ω =.065 π =.408 radians per minute. The linear velocity is therefore v = rω = 10(.408) 49 feet per minute. b) The angles are supplementary, so (3x 3 ) + (8x + 38 ) = 180. Solve for x to get x = 15 ; the angles are therefore 3(15 ) 3 = and 8(15 ) + 38 = 158. 6. a) The angles sum to 180 : x + 5 + 83 = 180. Solve for x to get x = 45. b) The angles are equal, so 3x 3 = x + 48. Solve for x to get x = 39. 6
.1. Fall 016 Exam 1 c) This shape has five sides, so its angles must add to 180 (5 ) = 540. Therefore x + x + 15 + 90 + 150 = 540. Solve for x to get x = 175 3 58.33. 7
.. Fall 016 Exam. Fall 016 Exam No calculator allowed on these problems 1. Throughout this problem, assume sin θ = 3. a) What is sin( θ)? b) If cos θ > 0, find cos θ. c) If cos θ > 0, what quadrant is θ in? d) What is sin(θ + 360 )? e) (Bonus) What is sin(θ + 180 )?. Find the exact value of each quantity: a) sin 45 b) cos 10 c) sin 180 d) cos 300 e) sin 570 3. Find the exact value of each quantity: f) cos( 30 ) a) sin 5π 6 b) sin 8π c) sin 3π 4 d) cos 10π 3 e) sin π f) cos 7π Calculator allowed on the rest of the exam 4. Find cos θ, if θ is as in the following picture: θ 19 7 5. Suppose sin θ =.55. Find all possible values of cos θ. 6. a) Find all angles θ between 0 and 360 such that sin θ =.3. b) Find all angles θ between 0 and 360 such that sin θ = 1.3. c) Find all angles θ between 0 and 360 such that cos θ =.75. d) Find all angles θ between 0 and 360 such that cos θ = 1. 7. Find the area of each indicated triangle: 8
.. Fall 016 Exam 9. 40 a) 1.5 15 0 b) 7 8. Solve triangle ABC, given the information in the picture below: A B 5 4.3 C 9. Solve triangle GHI if g = 17, h = 5 and I = 6. 10. Solve triangle ABC if a = 17, b = 10 and B = 40. 11. A person wants to measure the height of a flagpole. She walks 50 feet away from the base of the flagpole, turns around and notices that her angle of elevation to the top of the pole is 48. If her eyes are 5 feet above the ground, how tall is the flagpole? 9
.. Fall 016 Exam Exam Solutions 1. a) sin( θ) = sin θ = 3. b) Start with the Pythagorean identity: cos θ + sin θ = 1 ( ) cos θ + = 1 3 cos θ + 4 9 = 1 cos θ = 5 9 cos θ = ± 5 9. Since we are told cos θ > 0, cos θ = 5 9 = 5 3. c) Since sin θ < 0 and cos θ > 0, θ must be in Quadrant IV. d) sin(θ + 360 ) = sin θ = 3. e) θ + 180 is in Quadrant II, but has the same reference angle as θ, so sin(θ + 180 ) = 3.. a) sin 45 =. b) cos 10 = 1 (Quadrant II; reference angle 60 ) c) sin 180 = 0 (point on unit circle is ( 1, 0)) d) cos 300 = 1 (Quadrant IV; reference angle 60 ) e) sin 570 = sin 10 = 1 (Quadrant II; reference angle 30 ) f) cos( 30 ) = cos 30 = 3 3. a) sin 5π 6 = sin 150 = 1 (Quadrant II; reference angle 30 ) b) sin 8π = sin 0 = 0 c) sin 3π = sin( 135 ) = (Quadrant III; reference angle 45 ) 4 d) cos 10π = cos 4π = cos 3 3 40 = 1 (Quadrant III; reference angle 60 ) e) sin π = sin 90 = 1 (point on unit circle is (0, 1)) f) cos 7π = cos 7π = cos 70 = 0 (point on unit circle is (0, 1)) 4. First, use the Pythagorean Theorem to find the adjacent side, which I ll call a: a + 7 = 19 a = 19 7 = 31 17.66. Then cos θ = adjacent adjacent 17.66 19 =.997. 10
.. Fall 016 Exam 5. Start with the Pythagorean identity: cos θ + sin θ = 1 cos θ + (.55) = 1 cos θ +.305 = 1 cos θ =.6975 cos θ = ±.6975 ±.835. 6. a) From a calculator, θ = sin 1 (.3) 17.46. A second angle which solves the equation is 180 θ 16.54. b) There are no such angles (because 1.3 > 1). c) From a calculator, θ = cos 1 (.75) 41.4. A second angle which solves the equation is 360 θ = 318.6. d) The only such angle is θ = 0 (if you also included θ = 360, that s okay, but 360 is really the same angle as 0 ). 7. a) By the SAS Area Formula, A = 1 ab sin C = 1 (9.)(1.5) sin 40 36.96 sq units. b) Use Heron s Formula (so first, find the semiperimeter): s = a + b + c = 7 + 15 + 0 = 1. Then the area is A = s(s a)(s b)(s c) = 1(1 7)(1 15)(1 0) = 1(14)(6)(1) = 1764 8. First, the third angle is A = 90 5 = 65. = 4 sq units. Second, find the hypotenuse c using a trig function (either sine or cosine): sin 5 = 4.3 c.46 = 4.3 c.46c = 4.3 c = 4.3.46 = 10.175. 11
.. Fall 016 Exam Last, find the remaining side using the Pythagorean Theorem: a + b = c a + (4.3) = (10.175) a = (10.175) (4.3) = 85.03 9.. 9. The given information is SAS, so start with the Law of Cosines to find side i: i = g + h gh cos I i = 17 + 5 (17)(5) cos 6 i = 89 + 65 850(.4695) i = 514.949 i = 514.949.69. Now use the Law of Sines (or the Law of Cosines) again to find G (you could find H first instead): sin G = sin I g i sin G sin 6 = 17.69 sin G 17 =.889.69 sin G = 17(.889) =.6615.69 G = sin 1 (.6615) = 41.4. Last, find H: H = 180 6 41.4 = 76.6. 10. The given information is SSA, which is the ambiguous case of the Law of Sines: sin A = sin B a b sin A sin 40 = 17 10 10 sin A = 17(.647) sin A = 17(.647) 10 = 1.09. This equation has no solution (because 1.09 > 1), so there is no such triangle. 1
.. Fall 016 Exam 11. Draw a right triangle where the opposite side is the flagpole. We are given that the adjacent side is 50 feet and the angle to the horizontal is 48. That means, labelling the adjacent side as a, that tan 48 = opposite adjacent 1.11 = a 50 1.11(50) = a 55.5 = a. Adding the 5 feet to account for the height of the viewer s eyes, we see that the height of the flagpole is 55.5 + 5 = 60.5 feet. 13
.3. Fall 016 Exam 3.3 Fall 016 Exam 3 No calculators allowed on these problems 1. Throughout this problem, assume sec θ = 4. a) What is sec( θ)? b) What is cos θ? c) What is sec(θ 360 )? d) Which two quadrants might θ be in? e) If tan θ < 0, find cot θ.. Find the exact value of each quantity: a) sec 45 b) tan 150 c) cot 180 d) csc( 10 ) e) sec 60 3. Find the exact value of each quantity: a) tan 5π d) cos π b) csc π 6 c) tan 3π 4 e) sin π 3 4. Find the exact value of each quantity: a) sin 15 + cos 15 b) cot(45 + 45 ) c) tan 5π 6 d) 3 sin π e) sin 3 π Calculators allowed on the rest of this exam 5. In each diagram below, write an equation for x in terms of the other given quantities in the picture. θ x 17 w x a) θ b) 6. Suppose that sin θ =.614 and that tan θ > 0. Find the values of all six trig functions of θ. 7. Use a calculator to compute these quantities: 14
.3. Fall 016 Exam 3 a) csc 75 b) cot 40 + cot 70 c) tan 15 d) 3 sec 40 8. a) Find all angles θ between 0 and 360 such that tan θ =.7. b) Find all angles θ between 0 and 360 such that sec θ =.8. c) Find all angles θ between 0 and 360 such that csc θ =.35. 9. Throughout this question, assume that u = 5, 3, v = 1, 8 and that w is a vector whose magnitude is 7 and whose direction angle is 58. a) Compute u + v. b) Compute u v. c) Compute the angle between u and v. d) Compute the direction angle of v. e) Compute u + v. f) Write w in component form. 10. Let v and w be the vectors indicated in the picture below. a) Sketch the vector v + w on the picture; label that vector v + w. b) Sketch the vector w v on the picture; label that vector w v. w v 11. A bird leaves its nest and flies 0 miles on a heading 37, measured east of north. The bird then flies 9 miles on a bearing 7, measured west of north. a) How far is the bird from its nest? b) If the bird wanted to return to its nest, would it be more accurate to say it should fly southeast, or southwest? Explain your answer. 15
.3. Fall 016 Exam 3 Exam 3 Solutions 1. a) sec( θ) = sec θ = 4. b) cos θ = 1 sec θ = 1 4. c) sec(θ 360 ) = sec θ = 4. d) θ must be in Quadrant I or Quadrant IV. e) Since tan θ < 0, we now know θ is in Quadrant IV, so cot θ < 0. Sketch a triangle with hypotenuse 4 and adjacent side of length 1. From the Pythagorean Theorem, the opposite side is 4 1 = 15, so cot θ = adj opp = 1 15.. a) sec 45 =. b) tan 150 = 1 3 (reference angle 30, Quadrant II). c) cot 180 DNE. d) csc( 10 ) = 3 (reference angle 60, Quadrant III). e) sec 60 =. 3. a) tan 5π = 0 b) csc π 6 = c) tan 3π 4 = 1 (reference angle 45, Quadrant II) d) cos π = 0 e) sin π 3 = 3 (reference angle 60, Quadrant II) 4. a) sin 15 + cos 15 = 1 (by a Pythagorean identity) b) cot(45 + 45 ) = cot 90 = 0 c) tan 5π 6 = ( 1 3 ) = 1 3 d) 3 sin π = 3(1) = 3 e) sin 3 π = 1 5. a) x 17 = hyp opp b) x w = adj hyp = csc θ so x = 17 csc θ. = cos θ so x = w cos θ. 6. Since sin θ > 0 and tan θ > 0, θ is in Quadrant I, so all the trig functions are positive. We are given sin θ =.614, so csc θ = 1 = 1 = 1.69. From the sin θ.614 identity sin θ + cos θ = 1, we can solve for cos θ to get cos θ =.789. Then sec θ = 1 = 1 sin θ = 1.67. Last, tan θ = =.614 cos θ =.777 and cot θ = = cos θ.789 cos θ.789 sin θ.789 = 1.85..614 16
.3. Fall 016 Exam 3 7. a) csc 75 = 1.0358. b) cot 40 + cot 70 = 1.19175 +.36397 = 1.5557. c) tan 15 = ( 1.4815) =.03961. d) 3 sec 40 = 3(1.30541) = 3.916. 8. a) θ = tan 1.7 = 35.7539 ; a second angle is θ + 180 = 15.7539. b) If sec θ =.8, then cos θ = 1 so θ =.8 cos 1 ( 1 ) =.8 69.07. A second angle is 360 θ = 90.93. c) There are no such angles, because csc θ cannot be between 1 and 1. 9. Throughout this question, assume that u = 5, 3, v = 1, 8 and that w is a vector whose magnitude is 7 and whose direction angle is 58. a) u + v = 10, 6 + 1, 8 = 11, 14. b) u v = ( 5)( 1) + 3(8) = 5 + 4 = 9. c) First, u = ( 5) + 3 = 34 5.83. Second, v = 65 = 8.06. Therefore, ( 1) + 8 = u v = u v cos θ 9 = (5.83)(8.06) cos θ.6171 = cos θ 51.89 = θ. d) θ = tan 1 b = a tan 1 8 = 1 tan 1 ( 8) = 8.875. But this is in the wrong quadrant (v is in Quadrant II), so we need to add 180 to get 97.15. e) u + v = 6, 11 = ( 6) + 11 = 36 + 11 = 157 1.53. f) w = w cos θ, w sin θ = 7 cos 58, 7 sin 58 = 3.709, 5.936. 10.. v+w w-v w v 17
.3. Fall 016 Exam 3 11. a) Let v be the first part of the bird s journey; we have v = v cos θ, v sin θ = 0 cos 53, 0 sin 53 = 1.03, 15.97. Let w be the second part of the bird s journey; we have w = w cos θ, w sin θ = 9 cos 16, 9 sin 16 = 8.559,.781. Therefore the bird s position is v + w = 3.47, 18.75. The distance from the bird to the nest is v + w = 3.47, 18.75 = (3.47) + (18.75) = 19.06 mi. b) The first component of the bird s position is positive, so it has to fly southwest (although just barely west of south) to get back to its nest. 18
.4. Fall 016 Final Exam.4 Fall 016 Final Exam No calculators allowed on these questions 1. Find the exact value of each quantity: a) tan 45 b) cos 60 c) sin 90 d) sec 30 e) cos 0. Find the exact value of each quantity: a) cot 40 b) cos( 150 ) c) sin 315 d) csc 70 e) sec 450 3. Find the exact value of each quantity: a) sin 11π 6 b) sin π 3 c) cot 3π d) tan 5π 6 e) tan 11π 3 4. Find the exact value of each quantity: a) sin 50 + cos 50 c) sin 75 e) 8 sec π 4 b) cos 135 d) tan 4 30 5. Suppose cos θ = 3 and tan θ < 0. Find the exact values of all six trig functions 5 of θ. 6. Suppose tan θ = 3. a) Find tan θ. b) Find cot θ. c) Find cot θ. d) What two quadrants might θ be in? e) Find tan(θ + 70 ). f) Find tan( θ). 7. Sketch the graph of y = tan x. 8. a) Which of graphs A-K on the attached page (see page 1 of this pdf file) is the graph of y = sin x? b) Which of graphs A-K is the graph of y = sin x? c) Which of graphs A-K is the graph of y = sin x? d) Which of graphs A-K is the graph of y = sin(x + )? 19
.4. Fall 016 Final Exam e) Which of graphs A-K is the graph of y = sin x? f) Which of graphs A-K is the graph of y = sin x? 9. Vectors u, v and w are shown below. Sketch the following vectors on the same picture, being sure to label which vector is which: a) w u b) w c) 1 u d) u + v + w w v u 0
.4. Fall 016 Final Exam Graphs for Problem 8 Use these graphs to answer the questions in Problem 8: A B C 3 1 1 π π π - 1 - D E F 3 1 1 1 π -1 π π -1 4 π G H I 4-1 π -1 π π - - -3-3 J K 1 1-1 + π - π - -1 1
.4. Fall 016 Final Exam Calculators allowed on the rest of the exam 10. Evaluate the following expressions using a calculator. Your answers can (and should) be written as decimals. a) csc 5 b) sin 5 + sin 110 c) sec(83 40 ) d) tan 117 e) 4 cot 86 f) sin 100 cos 44 11. For each equation, find (decimal approximations of) all angles θ between 0 and 360 satisfying the following equations. If there are no such angles, say so. a) sin θ =.8 b) cos θ = 1.35 c) cot θ =.55 d) csc θ = 1 1. Suppose that θ is an angle in a right triangle whose hypotenuse has length 43. and whose side opposite θ has length 5.1. Find the six trig functions of angle θ. 13. a) Compute the norm of the vector 5, 4. b) Compute the direction angle of the vector 5, 7. c) Suppose v = 6, 4 and w =, 11. Compute 5v + w. d) Compute v w, where v and w are as in part (c). e) Find the components of a vector which has norm 17 and direction angle 30. 14. Solve triangle ABC, which is pictured below: C B 18 35 A
.4. Fall 016 Final Exam 15. Solve triangle EF G, which is pictured below at left: F 8 47 Q E 5 77 R G P 6 6.43 16. Solve triangle P QR, which is pictured above at right. 17. Classify the following statements as true or false: a) tan θ 1 = sec θ b) cos( θ) = cos θ c) cos(90 θ) = sin θ d) 1 cos θ = sin θ e) tan θ cos θ = sin θ 18. Answer any two of the following five questions. a) A wedge-shaped piece of pie taken from a pie with a 16 diameter has angle 50. Find the area of the piece of pie. b) A wheel of radius 18 feet rotates at a speed of 4 revolutions per minute. What is the linear velocity of a point on the edge of the wheel? c) Find the area of triangle ABC, if a = 30, b = 18 and C = 48. d) A 16-foot ladder leans up against the side of the building. If the top of the ladder is 13.5 feet above the ground, what angle does the ladder make with the ground? e) A boat leaves a harbor and sails 30 miles on a heading 40, measured east of north, then sails 0 miles due east. How far is it from the harbor? 3
.4. Fall 016 Final Exam Final Exam Solutions 1. a) tan 45 = 1 b) cos 60 = 1 c) sin 90 = 1 d) sec 30 = 1 cos 30 = 1 3/ = 3 e) cos 0 = 1. a) cot 40 = 1 3 (quadrant III, reference angle 60 ) b) cos( 150 ) = cos 150 = 3 (quadrant II, reference angle 30 ) c) sin 315 = (quadrant IV, reference angle 45 ) d) csc 70 = 1 = 1 (the point (0, 1) on the unit circle) 1 e) sec 450 = 1 cos 90 = 1 0 DNE. 3. a) sin 11π 6 = 1 (quadrant IV, reference angle 30 ) b) sin π = 3 (quadrant IV, reference angle 60 ) 3 c) cot 3π cos 70 = = 0 = 0 (the point (0, 1) on the unit circle) sin 70 1 d) tan 5π = 1 6 3 (quadrant II, reference angle 30 ) e) tan 11π 3 = 3 (quadrant IV, reference angle 60 ) 4. a) sin 50 + cos 50 = 1 b) cos 135 = ( ) = = 1. 4 c) Use the addition identity for sine: sin 75 = sin(30 + 45 ) = sin 30 cos 45 + cos 30 sin 45 = 1 3 + + 6 =. 4 d) tan 4 30 = tan(4 30 ) = tan 10 = 3. e) 8 sec π 4 = 8. 5. Since cos θ > 0 and tan θ < 0, θ must be in Quadrant IV. Therefore all the trig functions of θ are negative except for cosine and secant. Next, draw a triangle with adjacent side 3 and hypotenuse 5. Solve for the opposite side using the Pythagorean Theorem to get 5 3 = 5 9 = 16 = 4. Therefore cos θ = 3 5 ; sec θ = 5 3 ; sin θ = 4 5 ; csc θ = 5 4 ; tan θ = 4 3 ; cot θ = 3 4. 4
.4. Fall 016 Final Exam 6. a) tan θ = tan θ 1 tan θ = (3) 1 3 = 6 8 = 3 4. b) cot θ = 1 tan θ = 1 3. c) cot θ = 1 tan θ = 4 3. d) Since tan θ > 0, θ is in Quadrant I or III. e) tan(θ + 70 ) = tan θ = 3. f) tan( θ) = tan θ = 3. 7. The graph of y = tan x is on page 13 of my lecture notes. 8. a) This graph has period π = 4π, so it is graph F. 1/ b) This graph has been stretched vertically by a factor of, so it is graph A. 9. c) This graph is shifted down units, so it is graph G. d) This graph is shifted left units, so it is graph K. e) This graph has period π = π, so it is graph E. f) This graph has been squashed vertically to be half as tall, so it is graph B. u+v+w w-u u -w 10. a) csc 5 = 1 sin 5 = 1.690. b) sin 5 + sin 110 =.4618 +.939693 = 1.3631. c) sec(83 40 ) = sec 43 = 1 cos 43 = 1.36733. d) tan 117 = ( 1.9661) = 3.85184. e) 4 cot 86 = 4 ( ) 1 tan 86 = 4 1 = 4(.86745) = 1.14698. 3.48741 f) sin 100 cos 44 = (.984808)(.71934) =.708411. 5
.4. Fall 016 Final Exam 11. a) θ = sin 1.8 = 16.6 and 180 16.6 = 163.74. b) cos θ = 1.35 has no solution since 1.35 > 1. c) cot θ =.55 means tan θ = 1.55 = 1.81818 so θ = tan 1 ( 1.81818) = 118.81 and θ = 118.81 + 180 = 98.81. d) csc θ = 1 means sin θ = 1 1 = 1 so θ = sin 1 1 = 90. 1. First, the adjacent side is 43. 5.1 = 35.16. Therefore the six trig functions are sin θ = 5.1 43. =.581 csc θ = = 1.71 43. 5.1 tan θ = 5.1 35.16 =.713 cot θ = 35.16 5.1 = 1.4008 cos θ = 35.16 43. =.814 sec θ = = 1.9 43. 35.16 13. a) 5, 4 = 5 + ( 4) = 5 + 16 = 41. b) θ = tan 1 7 5 = tan.8 = 15.64. Since θ must be in Quadrant II, add 180 to get θ = 164.36. c) 5v + w = 30, 0 + 4, = 34,. d) Compute v w = 6() + ( 4)(11) = 1 44 = 3. e) 17 cos 30, 17 sin 30 = 10.97, 13.08. 14. B = 180 90 35 = 55. sin 35 = a 18 so a = 18 sin 35 = 10.3. cos 35 = b 18 so b = 18 cos 35 = 14.74. 15. Start with the Law of Cosines to find angle E: e = f + g fg cos E 47 = 5 + 8 (5)(8) cos E 09 = 65 + 784 1400 cos E 800 = 1400 cos E 800 1400 = cos E cos 1 8 14 = E 14.85 = E 6
.4. Fall 016 Final Exam Then use the Law of Cosines (or the Law of Sines) to find angle G: g = e + f ef cos G 8 = 5 + 47 (5)(47) cos G 784 = 65 + 09 350 cos G 050 = 350 cos G 050 350 = cos G 1 05 cos 35 = G 9.7 = G Last, angle F is 180 14.85 9.7 = 5.88. 16. First, angle Q is 180 6 77 = 77. Second, since angles Q and R are equal, sides q and r are equal so r = 6.43. Last, use the Law of Sines to find p: sin P = sin R p r sin 6 sin 77 = p 6.43.438371 =.97437 p 6.43.97437p = (.438371)6.43 p = (.438371)6.43.97437 =.89. 17. a) tan θ 1 = sec θ is FALSE (the tan and sec are backwards from what they are in the Pythagorean identity) b) cos( θ) = cos θ is FALSE (the sign disappears) c) cos(90 θ) = sin θ is TRUE (cofunction identity) d) 1 cos θ = sin θ is TRUE (Pythagorean identity) e) tan θ cos θ = sin θ is TRUE (write tan θ as sin θ and then cancel the cos θ cos θ terms on the left-hand side to see this) 18. a) The radius of the circle is r = 1 (16) = 8 inches; the angle in radians is 50 π =.87665. The area of the sector is therefore A = 1 180 r θ = 1 (8) (.87665) = 7.953 square inches. b) The angular velocity is ω = 4 pi = 8π = 5.137 rad/sec, so the linear velocity is v = rω = 18(5.137) = 45.389 feet per second. 7
.4. Fall 016 Final Exam c) By the SAS area formula, A = 1 ab sin C = 1 (30)18 sin 48 = 00.649 square units. d) The opposite side to the angle is 13.5 and the hypotenuse is 16, so the 1 13.5 angle is sin = 16 57.54. e) After the first heading, the boat is at position 30 cos 50, 30 sin 50 = 19.8,.98 (the angle is 50 because 90 40 = 50 ). After moving a further 0 miles east, the boat is at 19.8 + 0,.98 = 39.8,.98. The distance from the boat to the harbor is the norm of this vector, which is 39.8 +.98 = 45.50 miles. 8
Chapter 3 Exams from Fall 017 3.1 Fall 017 Exam 1 No calculator allowed on these problems 1. Convert the following angles from radians to degrees: a) 4π 3 b) 3π c) π. Draw the angle 50 in standard position. 3. Let sub be the function defined by sub x = x 3. Compute each quantity: a) sub 4 3 b) sub 4 3 c) sub (4) 3 d) 3 sub 4 Calculators allowed on the rest of the exam 4. Parts (a), (b), and (c) of this question are not related to one another. a) Find the measure of two different angles, one of which is less than 64 and one of which is greater than 64, both of which are coterminal with 64. b) Convert 5.6 radians to degrees (write your answer as an exact answer or as a decimal, rounded to two or more decimal places). c) Solve for x: x 100 3.5 = x 4.8 5. Parts (a), (b) and (c) of this question are not related to one another. 9
3.1. Fall 017 Exam 1 a) Find the distance between the points (1, 3) and (, 5) (write your answer either as an exact answer or as a decimal, rounded to two or more decimal places). b) The point (x, y) is on the unit circle. If x =.75, what are all possible values of y? c) Find the area of a sector whose central angle is 53, taken from a circle whose diameter is 6 feet. 6. A dragster has two front tires of radius 16 inches, and two rear tires of radius 6 inches. If the dragster is travelling with all four tires on the ground, with no skidding or friction, and rear tires rotate at 35 revolutions per second, what is the angular velocity of the front tires (in radians per second)? 7. In each problem (a)-(c), find x: a) Two complementary angles have measures x + 34 and 3x 0. b) The three angles of a triangle are x + 14, 3x 35 and 7x + 3. c) x is as in the picture below, where the vertical lines are parallel: 3x 5x -1 30
3.1. Fall 017 Exam 1 Solutions 1. a) 4π 3 = 4 π = 4 3 60 = 40. b) 3π = 3 π = 3 90 = 70. c) π = 180.. Since 50 is a little less than 70, this angle should have its terminal side just west of south (in Quadrant III): 3. a) sub 4 3 = sub(4) 3 = ((4) 3) 3 = 5 3 =. b) sub 4 3 = sub 1 = (1) 3 = 4 3 = 1. c) sub(4) 3 = ((4) 3) 3 = 5 3 = 15. d) 3 sub 4 = 3((4) 3) = 3(5) = 15. (This is the same as (b).) 4. a) For a smaller angle, subtract 360 from 64 to get 64 ; for a larger angle, add 360 to 64 to get 984. b) Multiply by 180 to get 30.85. π c) Cross-multiply to get 4.8(x 100) = 3.5x. Distribute on the left-hand side to get 4.8x 480 = 3.5x; combine the like terms to get 1.55x = 480; last, divide by 1.55 to get x = 309.677. 5. a) D = (x x 1 ) + (y y 1 ) = ( 1) + (5 3) = ( 3) + () = 9 + 4 = 13 3.6055. b) From the unit circle formula x + y = 1, we see x + (.75) = 1. Squaring out, this gives x +.0756 = 1, i.e. x = 1.0756 =.943. Take the square root of both sides to get x = ±.943 = ±.9614. c) The radius is half of the given diameter, i.e. r = 3. Convert the angle to radians by multiplying by π to get θ =.95 radians. Then A = 1 180 r θ = 1 (3) (.95) = 4.165 sq ft. 6. The rear tires have angular velocity ω rear = 35(π) = 19.911 rad/sec, so their linear velocity is v rear = r rear ω rear = 6(19.911) = 5717.1 in/sec. The front tires have the same linear velocity as the back tires, so v front = 5717.1 in/sec as well. Finally, ω front = v front r front = 5717.1 = 357.356 radians per second. 16 31
3.1. Fall 017 Exam 1 7. a) The angles are complementary, so (x + 34 ) + (3x 0 ) = 90. Combine the like terms to get 4x + 14 = 90, i.e. 4x = 76, i.e. x = 19. b) The angles sum to 180 : (x+14 )+(3x 35 )+(7x+3 ) = 180. Combine the like terms to get 11x 18 = 180, i.e. 11x = 198, i.e. x = 18. c) The angles are supplementary, so 3x + 5x 1 = 180. This means 8x = 19, i.e. x = 4. 3
3.. Fall 017 Exam 3. Fall 017 Exam No calculator allowed on these problems 1. Throughout this problem, assume cos θ = 3. 4 a) What is cos( θ)? b) What is cos(θ 360 )? c) What are the two quadrants θ could lie in? d) If sin θ < 0, find sin θ.. Find the exact value of each quantity: a) sin 180 c) sin 450 π e) sin 4 b) cos 60 d) cos 135 f) cos 3π g) sin 5π 6 h) cos π Calculators allowed on the rest of the exam 3. Find cos θ and sin θ, if θ is as in the following picture. (Either an exact answer or a decimal is OK, but make sure to tell me which answer is sin θ and which is cos θ.) θ 3 5 4. a) Find all angles θ between 0 and 360 such that sin θ =.65. b) Find all angles θ between 0 and 360 such that cos θ =. c) Find all angles θ between 0 and 360 such that cos θ =.43. 5. Find the area of this triangle: 37 55 15 1 6. A 0-foot long ladder leans up against the side of a building. If the bottom of the ladder makes an angle of 70 with the ground, how high up the side of the building does the ladder go? 33
3.. Fall 017 Exam 7. Solve triangle ABC, given the information in the picture below: B 43.5 A 38 C 8. Solve triangle KLM if k = 18, l = 13 and m = 10. 9. Solve triangle ABC if A = 70, B = 58 and a =. 34
3.. Fall 017 Exam Solutions 1. a) cos( θ) = cos θ = 3 4. b) cos(θ 360 ) = cos θ = 3 4. c) Since cos θ > 0, θ must be in Quadrants I or IV. d) Start with the Pythagorean identity: cos θ + sin θ = 1 ( ) 3 + sin θ = 1 4 9 16 + sin θ = 1 sin θ = 7 16 cos θ = ± 7 16. Since we are told sin θ < 0, sin θ = 7 16 = 7 4.. a) sin 180 = 0 (point on unit circle is ( 1, 0)). b) cos 60 = 1 c) sin 450 = sin 90 = sin 90 = 1 (point on unit circle is (0, 1)). d) cos 135 = (Quadrant II; reference angle 45 ) e) sin π = sin 4 45 = (Quadrant I) f) cos 3π = cos 540 = cos 180 = 1 (point on unit circle is ( 1, 0)) g) sin 5π 6 = sin( 150 ) = 1 (Quadrant III; reference angle π 6 = 30 ) h) cos π = cos 90 = 0 (point on unit circle is (0, 1)) 3. First, use the Pythagorean Theorem to find the hypotenuse, which I ll call c: Then cos θ = 3 + 5 = c c = 3 + 5 = 34 5.83. adjacent hypotenuse 3 5.83 =.5145 and sin θ = opposite hypotenuse 5 5.83 =.8575. 4. a) From a calculator, θ = sin 1 (.65) 15.37. Add 360 to this to get 344.63 ; take 180 the calculator answer to get the second answer which is 196.37. b) There are no such angles (because > 1). c) From a calculator, θ = cos 1 (.43) 64.5. A second angle which solves the equation is 360 θ = 95.5. 5. First, the third angle is 180 55 37 = 88. Then, by the SAS Area Formula, A = 1 ab sin C = 1 (1)(15) sin 88 89.94 sq units. 35
3.. Fall 017 Exam 6. In this problem, we have a right triangle where the hypotenuse is 0 and the angle from the hypotenuse to the ground is 70. We are asked to find the opposite side, which I ll call b. By SOHCAHTOA, we get sin 70 = opposite hypotenuse = b 0. Solve for b by cross-multiplying to get b = 0 sin 70 = 18.79 ft. 7. First, the third angle is B = 90 38 = 5. Second, find side a using a trig function (either sine or cosine): sin 38 = a 43.5.6157 = a 4 3.5 6.78 = a. Last, find the remaining side using the Pythagorean Theorem: a + b = c (6.78) + b = (43.5) b = (43.5) (6.78) = 1175 34.7. 8. The given information is SSS, so start with the Law of Cosines to find any angle (say K): k = l + m lm cos K 18 = 13 + 10 (13)(10) cos K 34 = 169 + 100 60 cos K 34 = 69 60 cos K 55 = 60 cos K.115 = cos K 10. = K Now use the Law of Cosines again (or the Law of Sines) to find a second 36
3.. Fall 017 Exam angle (say L): l = k + m km cos L 13 = 18 + 10 (18)(10) cos L 169 = 34 + 100 360 cos L 169 = 44 360 cos L 55 = 360 cos L.708 = cos L 44.9 = L Last, find the third angle (for me this is M): M = 180 10. 44.9 = 3.9. 9. The given information is SAA/AAS, which requires the Law of Sines. First, the third angle is C = 180 70 58 = 5. Second, find side length b using the Law of Sines: sin A = sin B a b sin 70 sin 58 = b b sin 70 = sin 58.939b = 18.65 b = 19.85. Last, find side length c using the Law of Sines (you could use the Law of Cosines): sin A = sin C a c sin 70 sin 5 = c c sin 70 = sin 5.939c = 17.33 c = 18.45 37
3.3. Fall 017 Exam 3 3.3 Fall 017 Exam 3 No calculator allowed on these problems 1. Throughout this problem, assume cot θ = 5. 3 a) What is cot( θ)? b) What is tan θ? c) Which two quadrants might θ be in? d) If sin θ < 0, find cos θ.. a) csc 135 b) tan 60 c) cot 180 d) sec( 5 ) e) cot 3π f) sec π 6 g) tan 3π 4 3. Find the exact value of each quantity: a) sec 4π 3 b) 8 sin π 6 c) sin 15 + cos 15 d) tan( 90 ) e) cos π + 1 4. In each diagram below, write an equation for x in terms of the other given quantities in the picture. Your formula for x should not contain division in it. x x 10 a) p θ b) α β 5. Let v and w be the vectors indicated in the picture below. a) Sketch the vector 1 w on the picture; label that vector 1 w. b) Sketch the vector v + w on the picture; label that vector v + w. v w 38
3.3. Fall 017 Exam 3 Calculators allowed on the rest of the exam 6. Suppose that θ is some angle such that when drawn in standard position, ( 7, 1) is on the terminal side of θ. Find sec θ and tan θ. 7. Use a calculator to compute decimal approximations of these quantities: a) csc 3 b) tan 140 + cot 140 c) cos 13 8. a) Find all angles θ between 0 and 360 such that tan θ = 4.15. b) Find all angles θ between 0 and 360 such that sec θ = 3.38. 9. a) Suppose u = 3, 8 and v = 4, 5. Find 3u + v. b) Find u v, where u and v are as in part (a). c) Suppose w is a vector with magnitude 7 and direction angle 85. Find the components of w. d) Find the magnitude of the vector z = 6.3, 5.. 10. Nemo the fish swims away from his home on a trajectory 50 north of east for miles, then changes course and swims due east for 1 mile. After that, Nemo changes course again and swims for 3 miles on a trajectory 35 south of east. At this point, what distance would Nemo have to swim if he wanted to swim directly home? 39
3.3. Fall 017 Exam 3 Solutions 1. a) cot( θ) = cot θ = 5 3. b) tan θ = 1 cot θ = 3 5. c) cot θ is positive in Quadrants I and III. d) Since cot θ > 0 and sin θ < 0, θ is in Quadrant III, so cos θ will be negative. Now draw a triangle and label the adjacent side as 5 and the opposite side as 3. Solve for the hypotenuse using the Pythagorean Theorem to get 5 + 3 = 34. Finally, cos θ =. a) csc 135 =. b) tan 60 = 3. c) cot 180 DNE. d) sec( 5 ) = sec 5 =. e) cot 3π = 0. f) sec π 6 = 3. g) tan 3π 4 = 1. 3. a) sec 4π 3 = ( ) = 4. b) 8 sin π 6 = 8 1 = 4. c) sin 15 + cos 15 = 1. d) tan( 90 ) = tan 180 = 0. e) cos π + 1 = 1 + 1 =. 4. a) x p = adjacent hypotenuse = cos θ, so x = p cos θ. adjacent = 5 hypotenuse 34. b) Label the vertical segment common to both triangles as w. Then w = 10 csc β and x = w csc α, so by substitution x = 10 csc β csc α. 5. To get 1 w, sketch a vector in the opposite direction as w, but half as long. To get v + w, first draw v (twice as long as v, in the same direction) and then add by drawing a parallelogram. You end up with the following: v + w 1 w 40
6. We have x = 7 and y = 1 so r = x + y = 193. Therefore 3.3. Fall 017 Exam 3 ( 7) + 1 = 49 + 144 = sec θ = r x = 193 7 1.984 and tan θ = y x = 1 7 1.714. 7. a) csc 3 = 1 sin 3 =.5593. b) tan 140 + cot 140 =.8391 + ( 1.1917) =.0308. c) cos 13 = (.8386) =.7033. 8. a) θ = tan 1 4.15 = 76.37. The other angle is 180 + θ = 56.37. b) cos θ = 1 = 1 =.958. Therefore θ = sec θ 3.38 cos 1.958 = 7.79. The other angle is 360 θ = 78.1. 9. a) 3u + v = 9, 4 + 4, 5 = 5, 9. b) u v = ( 3)4 + 8(5) = 8. c) w = 7 cos 85, 7 sin 85 = 1.811, 6.761. d) z = 6.3 + 5. = 66.73 = 8.16884. 10. Nemo s journey can be described by three vectors: the first part of his trip is the second part of his trip is the last part of his trip is a = cos 50, sin 50 = 1.85, 1.53 ; b = 1 cos 0, 1 sin 0 = 1, 0 ; c = 3 cos 35, 3 sin 35 =.457, 1.7. (The angle is 360 35 = 35 because the trajectory is south of east.) That means Nemo s eventual position is a + b + c = 1.85 + 1 +.457, 1.53 + 0 1.7 = 4.74,.188. The magnitude of this vector, which is the distance from Nemo to his home, is 4.74 + (.188) =.519 4.745 miles. 41
3.4. Fall 017 Final Exam 3.4 Fall 017 Final Exam No calculator allowed on these problems 1. Find the exact value of each quantity: a) cos 30 b) tan 90 c) sin 45 d) csc 90 e) sin 0. Find the exact value of each quantity: a) tan 150 b) cos( 5 ) c) sin 330 d) cot 180 e) sec 780 3. Find the exact value of each quantity: a) cos 4π 3 b) sin π 3 c) tan 3π 4 d) cot π 6 e) csc 5π 4. Find the exact value of each quantity: a) sin 310 + cos 310 b) cos 400 cos 40 c) tan 85 sec 85 d) sin 0 cos 40 + cos 0 sin 40 e) cos 75 5. Suppose tan θ = 3 of θ. 6. Suppose sin θ = 1 5 and sin θ < 0. Find the exact values of all six trig functions and cos θ > 0. a) Find csc θ. b) What quadrant is θ in? c) Find sin(θ 360 ). d) Find sin( θ). e) Find sin θ. f) Find csc θ. 7. Sketch a crude graph of each function, labelling the graph appropriately: a) y = sin x 4 b) y = 3 cos x c) y = cot x d) y = sin x + 1 8. Simplify the following expression using trig identities: 1 sin θ 1 csc θ 4
3.4. Fall 017 Final Exam Calculators allowed on the rest of the exam 9. Evaluate the following expressions using a calculator. Your answers can (and should) be written as decimals. a) sin 103 b) 5 cos( 51 ) c) cos 133 cos 75 d) csc(5 + 81 ) e) sin 31 f) tan 40 cot 108 g) sec 4 83 10. For each equation, find (decimal approximations of) all angles θ between 0 and 360 satisfying the following equations. If the equation has no solution, say so. a) sin θ =.683 b) cos θ =.3 c) tan θ = 3.1 d) sec θ = 0 e) cos θ =.735 f) sin θ = 1.04 11. Find the six trig functions of the angle θ, if the point (.7, 5.9) is on the terminal side of θ when it is drawn in standard position. 1. Let v = 8, 1 and let w = 3, 7. a) Compute the norm of v. b) Compute v w. c) Find the measure of the angle between v and w. 13. A chipmunk located 4 feet from a flagpole has to look up at an angle of 50 to see the top of the flagpole. How tall is the flagpole? (Assume the chipmunk s eyes are at ground level.) 14. A surveyor is trying to determine the distance between points X and Y as shown below, but can t measure the distance directly because there is an obstacle in the way. So instead, the surveyor measures the distances from X and Y to a third point Z, and measures an angle at point Z as shown in the 43
3.4. Fall 017 Final Exam figure below. Find the distance from point X to point Y. Y X obstacle 77 4 64 Z 15. Solve triangle P QR, where P = 75, Q = 58 and p = 53. 16. Choose any three of the following five questions. a) The video game Pac-Man features a character that looks like the lefthand figure below. Find the area enclosed by Pac-Man. 13 60 56 x.85 θ b) Write an expression for x in terms of the other quantities given in the right-hand figure above. Simplify your answer. c) A bicycle wheel of radius 16.5 inches rotates at a speed of 35 revolutions per minute. What is the linear velocity of a point on the edge of the wheel? d) Find the area of triangle ABC, if a = 6, b = 4 and c = 5. e) A bird leaves its nest and flies due west for 9 miles, then flies at an angle 57 north of east for 7 miles. How far from its nest is it? 44
3.4. Fall 017 Final Exam Solutions 1. a) cos 30 = 3. b) tan 90 = 1 which DNE. 0 c) sin 45 =. d) csc 90 = 1 1 = 1. e) sin 0 = 0.. a) tan 150 = 3 3 (quadrant II, reference angle 30 ) b) cos( 5 ) = (quadrant II, reference angle 45 ) c) sin 330 = 1 (quadrant IV, reference angle 30 ) d) cot 180 = 1 which DNE. 0 e) sec 780 = (quadrant I, reference angle 60 ) 3. a) cos 4π 3 = 1 (quadrant III, reference angle 60 ) b) sin π 3 = 3 (quadrant III, reference angle 60 ) c) tan 3π 4 = 1 (quadrant II, reference angle 45 ) d) cot π 6 = 3 (quadrant IV, reference angle 30 ) e) csc 5π = csc 90 = 1 1 = 1. 4. a) sin 310 + cos 310 = 1 (by the Pythagorean identity) b) cos 400 cos 40 = cos 40 cos 40 (by periodicity) which simplifies to 0. c) tan 85 sec 85 = 1 (by a rewritten version of tan θ + 1 = sec θ) d) Use the addition identity for sine to get sin 0 cos 40 + cos 0 sin 40 = sin(0 + 40 ) = sin 60 = 3 e) Use the addition identity for cosine to get cos 75 = cos(30 + 45 ) = cos 30 cos 45 sin 30 sin 45 = ( 3 ) ( ) ( 1 ) ( ) = 6 4. 5. Draw a triangle, labelling the opposite side and the adjacent side 3. Solve for the hypotenuse using the Pythagorean Theorem, obtaining 13. Since tan θ > 0 and sin θ < 0, the angle is in Quadrant III, so all trig functions other than tangent and cotangent are negative. Using the right triangle definitions of the trig functions, we get sin θ = 13 cos θ = 3 13 tan θ = 3 13 csc θ = 13 sec θ = 3 cot θ = 3 45
π π 3.4. Fall 017 Final Exam 6. a) csc θ = 1 sin θ = 1 1/5 = 5. b) Since sin θ and cos θ are positive, θ is in Quadrant I. c) sin(θ 360 ) sin θ = 1 5. d) sin( θ) = sin θ = 1. 5 e) First, find cos θ: draw a triangle, label the opposite side 1 and the hypotenuse 5, and solve for the adjacent side using the Pythagorean Theorem to get 4. Since the angle is in Quadrant I, cos θ = 4. Next, 5 sin θ = sin θ cos θ = ( 1 5 ) ( 4 5 f) csc θ = 1 sin θ = 1 4/5 = 5 4. ) = 4 5. 7. a) y = sin x 4 is the graph of y = sin x, shifted down 4 units: π -3-4 -5 b) y = 3 cos x is the graph of y = cos x, stretched vertically by a factor of 3: 3 π -3 c) y = cot x looks like this: d) y = sin x + 1 has period π B = π = π, and is shifted up 1 unit: 1 π π 46
3.4. Fall 017 Final Exam 8. Use Pythagorean identities, then a quotient identity, then divide the fractions: 9. a) sin 103 =.97437. 1 sin θ 1 csc θ = cos θ cot θ = cos θ cos θ sin θ b) 5 cos( 51 ) = 5(.693) = 3.1466. = cos θ 1 = sin θ 1 = sin θ. sin θ cos θ c) cos 133 cos 75 =.681998.58819 =.940817. d) csc(5 + 81 ) = csc 133 = 1 sin 133 = 1.731354 = 1.36733. e) sin 31 = (sin 31 ) = (.777146) =.603956. f) tan 40 cot 108 = (.8391) ( ) ( 1 tan 108 = (.8391) g) sec 4 83 = (sec(33 )) = ( ) ( 1 cos 33 = 1 3.07768 1.88948 ) =.764. ) =.6514. 10. a) θ = sin 1.683 = 43.1 ; the other angle is 180 θ = 136.9. b) θ = cos 1.3 = 76.7 ; the other angle is 360 θ = 83.3. c) θ = tan 1 3.1 = 7.7. Add 360 to get 87.3 ; the other angle is θ + 180 = 107.3. d) sec θ = 0 has solution because 1 < 0 < 1. e) θ = cos 1.735 = 137.3 ; the other angle is 360 θ =.7. f) sin θ = 1.04 has no solution because 1.04 > 1. 11. We are given x =.7, y = 5.9 and need to find r. By the Pythagorean Theorem, r = x + y = (.7) + 5.9 = 6.488. Then: sin θ = y r = 5.9 6.488 =.909 cos θ = x r =.7 6.488 =.416 tan θ = y x = 5.9.7 =.185 csc θ = r y = 6.488 5.9 = 1.099 sec θ = r x = 6.488.7 =.40 cot θ = x y =.7 5.9 =.416. 1. a) v = 8 + 1 = 65. b) v w = 8(3) + 1(7) = 4 + 7 = 31. 47
3.4. Fall 017 Final Exam c) Find the measure of the angle between v and w. Start with the formula v w = v w cos θ and solve for θ. From parts (a) and (b), we know v w and v, so we need to find w = 3 + 7 = 9 + 49 = 58. Now from the formula, we have v w = v w cos θ 31 = 65 58 cos θ 31 = cos θ 65 58.504883 = cos θ cos 1 (.504883) = θ 59.67 = θ. 13. Draw a right triangle, labelling the angle to the horizontal with 50. We are given that the adjacent side is 4 and need to find the opposite side, so use tangent: tan 50 = h 4 h = 4 tan 50 = 50.05 feet. 14. Use the Law of Cosines to find z, the distance between X and Y : z = x + y xy cos Z z = 4 + 77 (4)(77) cos 64 z = 1764 + 599 (4)(77)(.438371) z = 1764 + 599 835.38 z = 4857.6 z = 4857.6 = 69.6966. 15. First, the third angle is 180 75 58 = 47. Next, use the Law of Sines to find q: sin P = sin Q p q sin 75 sin 58 = 53 q.96596 =.848048 53 q.96596q = 53(.84808) = 44.9465 q = 44.9465.96596 = 46.531. 48
3.4. Fall 017 Final Exam Last, use the Law of Sines to find r: sin P = sin R p r sin 75 sin 47 = 53 r.96596 =.731354 53 q.96596q = 53(.731354) = 38.7617 q = 38.7617.96596 = 40.191. 16. a) The angle enclosed by Pac-Man is 360 56 = 304 = 304 π 180 = 5.3058 radians. Now by the area formula for sectors, A = 1 r θ = 1 (.85) (5.3058) = 1.80563 square units. b) Let w be the vertical line segment; using the right-hand triangle we have w = tan 13 60 = 3, so w = 13 3. Now from the left-hand triangle, we have x = csc θ so x = w csc θ. Substituting in for w, we get x = w 13 3 csc θ. c) ω = 35(π) = 19.911 radians per minute. Then the linear velocity is v = rω = 16.5(19.911) = 368.54 inches per minute. d) The semiperimeter is s = 1(a + b + c) = 1 (6 + 4 + 5) = 7.5; then by Heron s formula the area is A = s(s a)(s b)(s c) = 7.5(7.5 6)(7.5 4)(7.5 5) = 98.4375 = 9.9157 square units. e) The first part of the bird s journey is the vector v = 9, 0 ; the second part has angle 57 and magnitude 7 so its components are w = 7 cos 57, 7 sin 57 = 3.8147, 5.87069. Therefore the bird has travelled v + w = 5.18753, 5.87069 ; the distance from its nest is v + w = ( 5.18753) + 5.87069 = 61.3754 = 7.8345 miles. 49